I have found some solutions, but nothing general. The solutions are $f=g; f = g^{-1}; f(x)=x; f(x)=ax, g(x)=bx,$ for any $a$ and $b;$ I am unable to find any solutions other than these, or to prove that none exist.
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1Welcome to MSE! Your question would benefit from some additional context. For example, do your functions take real arguments? Are they real valued? – Reveillark Sep 25 '19 at 17:09
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2Have you heard about inverse of a function? – Vasili Sep 25 '19 at 17:16
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1that's one of the solutions I wrote in my description, sorry for not making it clear enough. If f and g are each other's inverses, then the condition is satisfied, but that's not the only solution to this functional equation. – Karen Mossoyan Sep 25 '19 at 17:23
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to clarify, the only restriction for the functions is that they have to take real values and output real values. – Karen Mossoyan Sep 25 '19 at 17:24
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2Well $f(x)=x^n$ and $g(x)=x^m$ seem to work. – Mark Bennet Sep 25 '19 at 17:27
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yeah, they work too, thanks for an answer! – Karen Mossoyan Sep 25 '19 at 17:32
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2$g=f\circ f\circ\cdots\circ f.$ – Thomas Andrews Sep 25 '19 at 17:36
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@ThomasAndrews that's much less trivial, than the ones I though up, thanks! – Karen Mossoyan Sep 25 '19 at 17:39
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You can also start with any $h$ and use $f=h\circ h\cdots \circ h$ $m$ times, and $g=h\circ h\circ\cdots \circ h$ $n$ times. Or, if $h$ is invertible, then $g=h^{-1}\circ h^{-1}\circ \cdots \circ h^{-1}$ $n$ times. – Thomas Andrews Sep 25 '19 at 18:12
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@ThomasAndrews that's a nice generalization – Karen Mossoyan Sep 25 '19 at 18:15